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Math 598 Feb 2, 20051
Geometry and Topology II

Spring 2005, PSU

Lecture Notes 5

2.2

Deﬁnition of Tangent Space

IfM is a smooth n-dimensional manifold, then to each point pofM we may
associate an n-dimensional vector spaceTpM which is deﬁned as follows. Let

CurvespM :={α: (−, )→M |α(0) =p}

be the space of smooth curves on M centered at p. We say that a pair of
curveα,β ∈CurvespM are tangent atp, and we writeα ∼β, provided that

there exists a local chart (U, φ) of M centered at psuch that
(φ◦α)(0) = (φ◦β)(0).

Note that if (V, ψ) is any other local chart of M centered atp, then, by the
chain rule,

(ψ◦α)(0) = (ψ◦φ−1◦φ◦α)(0)
=

(ψ◦φ−1)(φ(α(0))

◦(φ◦α)(0)

=

(ψ◦φ−1)(φ(β(0))

◦(φ◦β)(0)

= (ψ◦β)(0).

Thus∼is well-deﬁned, i.e., it is indpendent of the choice of local coordinates.
Further, one may easily check that ∼ is an equivalence relation. The set of
tangent vectors of M atp is deﬁned by

TpM :=CurvespM/∼.

Next we describe howTpM may be given the structure of a vector space.

Let (U, φ) denote, as always, a local chart ofM centered atp, and recall that

n = dim(M). Then we deﬁne a mapping f: TpM →Rn by

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Exercise 1. Show that the above mapping is well-deﬁned and is a bijection.
Since φ∗ is a bijection, we may use it to identify TpM with Rn and, in

particular, deﬁne a vector space structure on TpM. More explicitly, we set

[α] + [β] :=φ−∗1φ∗([α]) +φ∗([β]),

and

λ[α] :=φ−∗1λφ∗([α]).

2.3

Derivations

Here we give a more abstract, but useful, characterization for the tangent
space of a manifold, which reveals the intimate connection between tangent
vectors and directional derivatives.

LetC∞(M) denote the space of smooth functions on M and p∈M. We
say that two functions f, g ∈ C∞(M) have the same germ at p, and write

f ∼p g, provided that there exists an open neighborhood U of p such that

f|U =g|U. The reslting equivalence classes then deﬁnes the space of germ of

smooth functions of M atp:

Cp(M) :=C∞M/∼p .

Note that we can add and multiply the elements of CpM in an obvious way,

and with respect to these operations one may easily check that Cp(M) is an

algebra over the ﬁeld of real numbersR.

We say that a mappingD: Cp(M)→R is a derivation provided that D

is linear and satisﬁes the Leibnitz rule, i.e.,

D(f g) = Df·g(p) +f(p)·Dg

for all f, g ∈ Cp(M). If D1 and D2 are a pair of such derivations, then we

deﬁne their sum by (D1+D2)f :=D1f+D2f, and for anyλ∈R, the scalar

product is given (λD)f :=λ(Df).

Exercise 2. Show that the set of derivations of CpM forms a vector space

with respect to the operations deﬁned above.

Note that each element X ∈ TpM gives rise to a derivation of Cp(M) if,

for any f ∈Cp(M), we set

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where αX: (−, ) → M is a curve which belongs to the equivalence class

denoted by X, i.e., X = [αX].

Exercise 3. Check thatXf is well-deﬁned and is indeed a derivation.
A much less obvious fact, whose demonstrationis the main aim of this
section, is that, conversely, every derivation of Cp(M) corresponds to (the

directional derivative determined by) a tangent vecor. More formally, ifDpM

denotes the space of derivations of CpM, then

Theorem 4. TpM is isomorphic to DpM.

The rest of this section is devoted to the proof of the above result. To this
end we need a pair of lemmas. Let 0∈Cp(M) denote the constant function

zero, i.e. 0(p) := 0.

Lemma 5. If f ∈CpM is a constant function, then Df =0, for any D ∈

DpM.

Proof. First note that, since f is constant, say f(p) = λ,

D(f) =D(f ·1) =D(λ·1) =λD(1),

where 1 denotes the constant fucntion 1(p) = 1. Further,

D(1) = D(1·1) =D(1)·1 + 1·D(1) = 2D(1).

Thus D(1) = 0, which in turn yields that D(f) = 0.

Lemma 6. Let f: Rn → R be a smooth function. Then, for any p ∈ Rn,

there exist smooth functions gi: Rn→R, i= 1, . . . , n, such that

gi(p) = ∂f

∂xi

(p),

and

f(x) =f(p) +

n

i=1

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Proof. The fundamental theorem of calculus followed by chain rule implies
that

f(x)−f(p) =
1

d

dtf(tp+ (1−t)x)dt

=
1
0
n

i=1
∂f

dxi

tp+(1−t)x(x

i −pi)dt

=
n

i=1
1
0
∂f
dxi

tp+(1−t)xdt(x
i−

pi).

So we set

gi(x) :=

1
0

∂f
dxi

tp+(1−t)xdt.

Now we are ready to prove the main result of this section

Proof of Theorem 4. Recall that if (U, φ) is a local chart of M centered atp,
then the mapping [α]→(φ◦α)(0) is an isomorphism betweenTpM and Rn.

Similarly, f → f ◦φ−1 is an isomorphism between C

pM and CoRn, which

yields that DpM is isomorphic to DoRn. So it remains to show that DoRn

is isomorphic to Rn.

Letxi:Rn →R, given byxi(p) :=pi, be the coordinate functions of Rn.

It is easy to check that the mapping

DoRnD
F

−→(Dx1, . . . , Dxn)∈Rn

is a homomorphism. Furhter, F is one-to-one because, by the previous

lem-mas,

Df = 0 +

n

i=1

(Dgi·xi(o) +gi(o)·Dxi) =

n

i=1

∂f
∂xi

(o)Dxi.

In particular, knowledge ofDxi uniquely determinesD. Finally it remains to

show that F is onto. To this end note that to each X = (X1, . . . , Xn)∈Rn,

we may assign a derivation of CpRn given by

DX :=

n

i=1

Xi ∂
∂xi

x=o.

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Exercise 7. Show that any local chart (U, φ) ofM centered atpdetermines
a basis E1φ, . . . Eφ

n for TpM as follows. For every f ∈CpM, set:

Eiφf := ∂(f◦φ
−1)

∂xi

(o).

2.4

The diﬀerential map

Let f: M →N be a smooth map, and p∈M. Then the diﬀerential of f at

p is the mapping dfp: TpM →Tf(p)N given by

dfp([α]) := [f ◦α].

Exercise 8. Show that if f: Rn → Rm and we identify T

pRn and Tf(p)Rm

withRnandRmrespectively in the standard way (i.e., via the mapping [α]→

α(0)) then dfp may be identiﬁed with the linear transformation determined

by the jacobian matrix (∂fi/∂x

j) (in particular,dfp is a generalization of the

standard derivative Df(p) of maps between Euclidean spaces).

Using the characterization of TpM as the space of derivations over the

germ of smooth functions of M at p, one may give an alternative deﬁnition
of dfp as follows. Given X ∈TpM, we deﬁne

dfp(X) g :=X(g◦f),

for any g ∈Cf(p)N. Thus dfp(X)∈Df(p)N Tf(p)N. Note that if X = [α],

then

X(g◦f) = (g◦f◦α)(0) = [f◦α]g.

Thus the two deﬁnitions of dfp presented above are indeed equivalent. Using

the second deﬁnition, one may immediately check that dfp is a

homomor-phism. Another fundamental property is:

Exercise 9 (The chain rule). Show that iff: M →N andg: N →Lare
smooth maps, then, for any p∈M,

d(g◦f)p =dgf(p)◦dfp.

We say f: M → N is a diﬀeomorphism if f is a homeomorphism, and

f and f−1 are smooth. If there exists a diﬀeomorphism between a pair of

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Exercise 10. Show that if f: M → N is a diﬀeomorphism, then dfp is an

isomorphism for all p ∈ M. In particular, conclude that if M and N are
diﬀeomorphic, then dim(M) = dim(N).

Note that the last statement if the above exercise also follows from the
standard fact in Algebraic topology thatRn and Rm are homemorphic only

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Sach toan Tieng Anh hay 5G2010

<b>Lecture Notes 5</b>

<b>2.2</b>

<b>Deﬁnition of Tangent Space</b>

<b>2.3</b>

<b>Derivations</b>

<b>2.4</b>

<b>The diﬀerential map</b>

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